如何在R中传递...参数时保留!!运算符不被破坏？

Let's break down how to solve this problem, since match.call() doesn't play nicely with rlang's injection operators like !!.

The Issue with match.call() #

When you use match.call(expand.dots = FALSE)$..., it captures the raw parse tree of your function call. That's why !!my_variable gets turned into !(!my_variable) — base R's match.call() just sees two separate negation operators instead of the injection syntax that rlang uses for tidy evaluation.

Solution 1: Keep the !! Expression Intact (Your "期望行为") #

If you want to preserve the !!my_variable syntax in the output, use rlang::enquos() to capture the quoted arguments. This function is built to understand rlang's injection markers, so it won't break the !! syntax:

library(rlang)

f <- function(...){
  # Capture arguments as quosures to preserve injection syntax
  args <- enquos(...)
  # Extract just the expression part from each quosure
  lapply(args, quo_get_expr)
}

# Test it
my_variable <- "my value"
f(a = 1234, b = !!my_variable)
#> $a
#> [1] 1234
#> 
#> $b
#> !!my_variable

Solution 2: Resolve the Injected Value (Your "更优期望") #

If you want !!my_variable to expand to its actual value (so b returns "my value"), use rlang::list2() — it automatically handles injection in ... arguments:

f <- function(...){
  list2(...)
}

my_variable <- "my value"
f(a = 1234, b = !!my_variable)
#> $a
#> [1] 1234
#> 
#> $b
#> [1] "my value"

For more control, you can also capture quosures and evaluate them explicitly with eval_tidy():

f <- function(...){
  args <- enquos(...)
  lapply(args, eval_tidy)
}

f(a = 1234, b = !!my_variable)
#> $a
#> [1] 1234
#> 
#> $b
#> [1] "my value"

The core idea here is that base R tools like match.call() don't recognize rlang's tidy evaluation syntax. Using rlang's purpose-built functions lets you properly handle injection operators in ... arguments without breaking their intended behavior.

内容的提问来源于stack exchange，提问作者landau